Question

need help with derivative!

Answer 1

x = -5cosx and y=-5sinx

Answer 2

is dx = 5sinx and dy=-5cosx?

Answer 3

just want to make sure my answer is right

Answer 4

and how would you find d^2y/dx^2 of this?

Answer 5

would it be 5sin^(2)x + 5cos^(2)x / 5sin^(3)x?

Answer 6

`y=-5sinx`
I wouldn't notate the derivative of y, as dy.
Rather, I would say y' or f'(x).
but, yes the derivative of -5sin(x) is -5cos(x).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`x = -5cosx`
the derivative of x, is 1.
So, I would say ,
`1=5sin(x)`

Answer 7

wow.... first, I would like you to verify the initial question.

Answer 8

and the second deriv. of which function are looking for?

Answer 9

its asking for both

Answer 10

for both functions combined, or separately?

Answer 11

the x=-5cosx and y=-5sinx

Answer 12

and are you sure they are
x = -5cosx and y=-5sinx
and NOT
y = -5cosx and y=-5sinx

??

Answer 13

no its x and y value

Answer 14

like its asking for the slope and concavity but those answers come from the first derivative and second derivative

Answer 15

\(\large\color{black}{ \displaystyle x= -5\cos(x) }\) isn't really a function

Answer 16

its -5cos(theta)

Answer 17

i just typed x to make it easier

Answer 18

or is it
\(\large\color{black}{ \displaystyle y=-5\cos(\theta) }\) ?

Answer 19

the first function is x=-5cos(theta) and the second function is y=-5sin(theta)

Answer 20

so, if you have just -5cos(theta), well then write it as a function y=f(theta)
(which is what I did)
and then find it like that

Answer 21

\(\large\color{black}{ \displaystyle y=-5\cos(\theta) }\)
you were correct,
\(\large\color{black}{ \displaystyle y'=5\sin(\theta) }\)

Answer 22

I see what they are doing...

Answer 23

x is the output variable in the first one

Answer 24

so i would have dy/dx = 5sin(theta)/-5cos(theta)?

Answer 25

k, lets restart.....
functions: \(\large\color{black}{ \displaystyle x=-5\cos(\theta) }\) \(\large\color{black}{ \displaystyle y=-5\sin(\theta) }\)

you want: \(\large\color{black}{ \displaystyle dx/d\theta \\[0.9em] }\) \(\large\color{black}{ \displaystyle dy/d\theta }\)
and \(\large\color{black}{ \displaystyle d^2x/d\theta^2}\) \(\large\color{black}{ \displaystyle d^2y/d\theta^2 }\)

Answer 26

is this correct?

Answer 27

yes

Answer 28

yes

Answer 29

`x = -5cosx` and `y=-5sinx`

k, so for convenience will put the first derivatives using the "prime" notation
called the Leibniz.
\(\large\color{black}{ \displaystyle x'=5\sin(\theta) }\) \(\large\color{black}{ \displaystyle y'=5\cos(\theta) }\)

Answer 30

you got them correctly.

Answer 31

now you need x`` (using "x" here, like we normally use "y")
and for second one you need the y`` for second function.

Answer 32

do you know how to take the derivative of the derivative again?
(applying same techniques)

Answer 33

\(\large\color{black}{ \displaystyle x''=~? }\)
\(\large\color{black}{ \displaystyle y''=~? }\)

Answer 34

yeah d^2y/dx^2

Answer 35

no

Answer 36

you are mixing up the veriables

Answer 37

but thats what its asking me to find

Answer 38

`d^2x / d(theta)^2`
and the second one is
`d^2y / d(theta)^2`

Answer 39

they are asking you to find d^2y/dx^2 ?!

Answer 40

yeah

Answer 41

if possible

Answer 42

x = -5cos(theta) and y=-5sin(theta)

and find the d^2y / dx^2 for both functions ?

Answer 43

yes

Answer 44

like the question is asking to find dy/dx which i already got and d^2y/dx^2 if possible, and find the slope which i already got and concavity (if possible) at the point corresponding to theta=pi/4

Answer 45

oh, I see

Answer 46

they are using an awkward notation... but really you are looking for
d^2x / d(theta)^2
and
d^2y / d(theta)^2

but, they wanted to tell you to "find the second derivative", but posting d^2y/dx^2 (which is generally a notation, but incorrect notation-wise in this case).

Answer 47

\(\large\color{black}{ \displaystyle x=-5\cos(\theta) ~~~~~~~~~~~~~~~ y=-5\sin(\theta) }\)

so far we have:
\(\large\color{black}{ \displaystyle x'=5\sin(\theta) ~~~~~~~~~~~~~~~ y'=-5\cos(\theta) }\)

Answer 48

i thought it was taking the derivative with respect to x or something like that

Answer 49

yes

Answer 50

no, they just used that (faulty, and technically wrong for these cases) notation to tell you to find the second derivative instead of writing the words "second derivative".

Answer 51

it wouldn't make sense to find the second derivative of function y, with respect to x.

Why? Because there is not such a function y=f(x) .

Answer 52

Anyway, so you got
\(\large\color{black}{ \displaystyle x'=5\sin(\theta) ~~~~~~~~~~~~~~~ y'=-5\cos(\theta) }\)

and need,
\(\large\color{black}{ \displaystyle x''=~\text{______}~~~~~~~~~~~~~~~ y''=~ \text{______}}\)

Answer 53

x''=5cos(theta) y''=5sin(theta)

Answer 54

yes

Answer 55

done correctly

Answer 56

and so how do i uses this to find the concavity? do i divide dy''/dx''

Answer 57

lets do the concav. of the first function right now. K?

Answer 58

okay

Answer 59

are you given some interval ?

Answer 60

no just that the concavity is at point corresponding to theta=pi/4

Answer 61

and you don't need to know whether functions are in/de creasing at tht point ?

Answer 62

no

Answer 63

I disconnected, apologize

Answer 64

its okay

Answer 65

for the first function.
\(\large\color{black}{ \displaystyle \frac{d^2x}{d\theta^2}=5\cos\theta }\)

set it =0

\(\large\color{black}{ \displaystyle 0=5\cos\theta }\)

and wherever the 2nd deriv. is negative function is concave down, and wherever the 2nd deriv. is positive the function is concave up.

so lets plug in our value (i.e. pi/4) and see if the 2nd deriv. is pos. or neg. at this point.

\(\large\color{black}{ \displaystyle 5\cos\left( \pi/4\right) }\)

Answer 66

is \(\large\color{black}{ \displaystyle 5\cos\left( \pi/4\right) }\) `positive` (and then function is `concave up` at that point) OR `negative` (and then function is `concave down` at that point) ?

Answer 67

its positive 5sqrt(2)/2

Answer 68

yes, exactly

Answer 69

so concave up or down (at that point) ?

Answer 70

concave up

Answer 71

yup

Answer 72

now, lets perform the same thing, but with the second function.

Answer 73

k

Answer 74

it would be positve 5sqrt(2)/2 concave up

Answer 75

yes

Answer 76

sin(pi/4) = cos(pi/4)

we know this from a 45-45-90 degree triangle